Sure, let's determine the number of possible imaginary (complex) solutions for each equation by considering their degrees and the properties of their coefficients.
### Equation a) [tex]\( x^3 - x^2 - 36x + 36 \)[/tex]
This is a cubic equation (degree 3). According to the Fundamental Theorem of Algebra, a cubic equation with real coefficients has either:
- One real root and a pair of complex conjugate roots
- Three real roots
If there is a pair of complex solutions, they must be complex conjugates, thus:
- Possible imaginary solutions: 2
### Equation b) [tex]\( -2x + 2 \)[/tex]
This is a linear equation (degree 1). A linear equation has exactly one real root:
- Possible imaginary solutions: 0
### Equation c) [tex]\( x^5 - 43x^3 + 6x^2 + 252x - 216 \)[/tex]
This is a quintic equation (degree 5). According to the Fundamental Theorem of Algebra, it can have:
- One real root and two pairs of complex conjugate roots
- Three real roots and one pair of complex conjugate roots
- Five real roots
For cases involving complex roots, they come in pairs, hence:
- Possible imaginary solutions: 0, 2, 4
### Equation d) [tex]\( x^2 - 7x + 6 \)[/tex]
This is a quadratic equation (degree 2). A quadratic equation can have:
- Two real roots
- One real root (repeated)
- Two complex conjugate roots
If complex solutions exist, they occur in conjugate pairs. Therefore,
- Possible imaginary solutions: 0, 2
### Equation e) [tex]\( x^4 - 3x^3 - 34x^2 + 108x - 72 \)[/tex]
This is a quartic equation (degree 4). According to the Fundamental Theorem of Algebra, it can have:
- Four real roots
- Two pairs of complex conjugate roots
- Two real roots and one pair of complex conjugate roots
For cases involving complex roots, they come in pairs, hence:
- Possible imaginary solutions: 0, 2, 4
In conclusion, the number of possible imaginary solutions for each equation is:
- (a) 2
- (b) 0
- (c) 0, 2, 4
- (d) 0, 2
- (e) 0, 2, 4
